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Question
Prove that:
`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`
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Solution
L.H.S. = `(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB)`
= `((sinA - sinB)(sinA + sinB) + (cosA - cosB)(cosA + cosB))/((cosA + cosB)(sinA + sinB)`
= `(sin^2A - sin^2B + cos^2A - cos^2B)/((cosA + cosB)(sinA + sinB))`
= `((sin^2A + cos^2A) - (sin^2B + cos^2B))/((cosA + cosB)(sinA + sinB))` ...[∵ cos2 A + sin2 A = 1]
= `(1 - 1)/((cosA + cosB)(sinA + sinB))`
= 0 = R.H.S.
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